Question: $f(r)=r^2+4r-60$ 1) What are the zeros of the function? Write the smaller $r$ first, and the larger $r$ second. $\text{smaller }r=$
To find the zeros of the function, we need to solve the equation $f(r)=0$. We can do that by factoring $f(r)$. $\begin{aligned} r^2+4r-60&=0 \\\\ (r+10)(r-6)&=0 \\\\ r+10=0&\text{ or }r-6=0 \\\\ r={-10}&\text{ or }r={6} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $r$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }r\text{-coordinate}&=\dfrac{({-10})+({6})}{2} \\\\ &={-2} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $f({-2})$ : $\begin{aligned} f({-2})&=({-2})^2+4({-2})-60 \\\\ &=4-8-60 \\\\ &=-64 \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }r&=-10 \\\\ \text{larger }r&=6 \end{aligned}$ The vertex of the parabola is at $(-2,-64)$